Problem 65 - Algebra and Trigonometry

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\sec t=3, \quad\) terminal point of \(t\) is in quadrant IV

Step-by-Step Solution

Answer

\( \sin t = -\frac{2\sqrt{2}}{3}, \tan t = -2\sqrt{2}, \csc t = -\frac{3\sqrt{2}}{4}, \cot t = -\frac{\sqrt{2}}{4} \).

Step 1: Recall the definition of secant

The secant of an angle is the reciprocal of the cosine of that angle. Therefore, if \( \sec t = 3 \), it implies that \( \cos t = \frac{1}{3} \).

Step 2: Determine the cosine sign based on quadrant

Since the terminal point of \( t \) is in the fourth quadrant, \( \cos t \) is positive. Thus \( \cos t = \frac{1}{3} \).

Step 3: Find the sine using Pythagorean identity

Use the identity \( \sin^2 t + \cos^2 t = 1 \) to find \( \sin t \). Substituting \( \cos t = \frac{1}{3} \), we get \( \sin^2 t = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \). Then, \( \sin t = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3} \) because sine is negative in the fourth quadrant.

Step 4: Calculate tangent from sine and cosine

Tangent is defined as \( \tan t = \frac{\sin t}{\cos t} \). Substitute the values found: \( \tan t = \frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = -2\sqrt{2} \).

Step 5: Find the remaining trigonometric functions

From the values of sine, cosine, and tangent, we find the other trigonometric functions: \( \csc t = \frac{1}{\sin t} = -\frac{3}{2\sqrt{2}} = -\frac{3\sqrt{2}}{4} \), and \( \cot t = \frac{1}{\tan t} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4} \).

Key Concepts

Secant

The secant function is one of the six primary trigonometric functions and relates closely to cosine. In fact, secant is essentially the flip or reciprocal of cosine. This means whenever you see secant, you're looking at:
\[\sec t = \frac{1}{\cos t}\]
This relationship is particularly useful because it helps us find the cosine value if we know the secant, like in our problem where \( \sec t = 3 \). By taking the reciprocal, we find that \( \cos t = \frac{1}{3} \). Understanding this relationship simplifies solving many trigonometric problems.

• Remember: You can find the cosine of an angle by simply taking the reciprocal of the given secant value.
• Secant, like other trigonometric functions, can provide both positive and negative values depending on the quadrant in which the angle lies.

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry that relates sine and cosine values on the unit circle. It is written as:
\[\sin^2 t + \cos^2 t = 1\]
This identity is powerful because it allows us to find one trigonometric function if we know the other. In our exercise, knowing the cosine value \( \cos t = \frac{1}{3} \), we can use the identity to solve for sine. By rearranging the formula:
\[\sin^2 t = 1 - \cos^2 t\]Substitute \( \cos t = \frac{1}{3} \):
\[\sin^2 t = 1 - \left(\frac{1}{3}\right)^2 = \frac{8}{9}\]Then, taking the square root, we find \( \sin t = \pm \frac{2\sqrt{2}}{3} \). The Pythagorean identity assists in deducing relationships between trigonometric identities in different quadrants, emphasizing its utility in solving complex trigonometric problems.

Quadrant IV

In trigonometry, the coordinate plane is divided into four quadrants, and each quadrant determines the sign of trigonometric functions. Quadrant IV is located in the lower right section of a typical coordinate plane. In this quadrant:

• Cosine (\( \cos t \)) is positive.
• Sine (\( \sin t \)) is negative.
• Tangent (\( \tan t \)) is negative (since it is the ratio of a negative sine and a positive cosine).

Understanding the signs of trigonometric functions in each quadrant is crucial for solving problems. Our problem confirms that since the terminal point lies in Quadrant IV, the cosine remains positive while sine and tangent become negative. This takes away the guesswork when determining function values based on trigonometric identities and known conditions.